Derivative of the Lorentz factor

[SiennaTheGr8]

According to David Morin (link: https://books.google.com/books?id=Ni6CD7K2X4MC&pg=PA636), the time-derivative of the Lorentz factor is ($c=1$):

$\dot{\gamma} = \gamma^3 v \dot{v}$,

and the four-acceleration:

$\mathbf{A} = (\gamma^4 v \dot{v}, \gamma^4 v \dot{v} \mathbf{v} + \gamma^2 \mathbf{a})$.

Is that a mistake? I thought that it's:

$\dot{\gamma} = \gamma^3 (\mathbf{v} \cdot \dot{\mathbf{v}})$,

and

$\mathbf{A} = (\gamma^4 (\mathbf{v} \cdot \dot{\mathbf{v}}), \, \gamma^4 (\mathbf{v} \cdot \dot{\mathbf{v}}) \mathbf{v} + \gamma^2 \mathbf{a})$.I could be wrong, but my understanding is that the $v^2$ in the radicand is really $\mathbf{v} \cdot \mathbf{v}$, so that:

$\dfrac{d}{dt} \, (1 - v^2)^{-1/2} = \dfrac{d}{d(v^2)} \, (1 - v^2)^{-1/2} \, \dfrac{d}{dt} \, (\mathbf{v} \cdot \mathbf{v})$.

 

[Orodruin]

Since $v^2 = \vec v \cdot \vec v$, it follows directly that $v \dot v = \vec v \cdot \dot{\vec v}$. However, note that $\dot v$ is generally not equal to $|\dot{\vec v}|$.